Topology and its Applications 160 (2013) 553–563
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Topology and its Applications www.elsevier.com/locate/topol
A Suzuki type fixed point theorem for a generalized multivalued mapping on partial Hausdorff metric spaces Mujahid Abbas a,b , Basit Ali b , Calogero Vetro c,∗ a b c
Department of Mathematics and Applied Mathematics, University of Pretoria, Lynnwood road, Pretoria 0002, South Africa Department of Mathematics, Lahore University of Management Sciences, 54792 Lahore, Pakistan Dipartimento di Matematica e Informatica, Università degli Studi di Palermo, Via archirafi 34, 90123 Palermo, Italy
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 10 October 2012 Received in revised form 1 December 2012 Accepted 4 January 2013
In this paper, we obtain a Suzuki type fixed point theorem for a generalized multivalued mapping on a partial Hausdorff metric space. As a consequence of the presented results, we discuss the existence and uniqueness of the bounded solution of a functional equation arising in dynamic programming. © 2013 Elsevier B.V. All rights reserved.
MSC: 47H09 47H10 54H25 Keywords: Common fixed point Multi-valued mapping Partial metric space
1. Introduction and preliminaries In 1937, Von Neumann [34] initiated the fixed point theory for multivalued mappings in the study of game theory. Indeed, the fixed point theorems for multivalued mappings are quite useful in control theory and have been frequently used in solving many problems of economics and game theory. Successively, Nadler [24] initiated the development of the geometric fixed point theory for multivalued mappings. He used the concept of the Hausdorff metric to establish the multivalued contraction principle containing the Banach contraction principle as a special case. Here, we recall that a Hausdorff metric H induced by a metric d on a set X is given by
H ( A , B ) = max sup d(x, B ), sup d( y , A ) x∈ A
y∈ B
for every A , B ∈ C B ( X ), where d(x, B ) = inf{d(x, y ): y ∈ B } and C B ( X ) is the collection of the closed and bounded subsets of X . Theorem 1.1. ([24]) Let ( X , d) be a complete metric space and T : X → C B ( X ) be a multivalued mapping satisfying
H ( T x, T y ) kd(x, y ) for all x, y ∈ X and k ∈ (0, 1). Then T has a fixed point, that is, there exists a point z ∈ X such that z ∈ T z.
*
Corresponding author. E-mail addresses:
[email protected] (M. Abbas),
[email protected] (B. Ali),
[email protected] (C. Vetro).
0166-8641/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.topol.2013.01.006
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In the last decades, a number of fixed point results (see, for example, [10,11,13,14,20–22,27,31]) have been obtained in attempts to generalize Theorem 1.1. One of the most significant fixed point theorems for multivalued contractions appeared in [18, Theorem 2.1]. To state this theorem, we need the following notation and notion. Let ( X , d) be a metric space and T : X → C B ( X ) be a multivalued mapping. Define
M d (x, y ) = max d(x, y ), d(x, T x), d( y , T y ),
d(x, T y ) + d( y , T x)
2
(1)
for all x, y ∈ X . Also, let ψ : [0, 1) → (0, 1] be the non-increasing function defined by
ψ(r ) =
1 1−r
if 0 r < 12 , if
1 2
(2)
r < 1.
Then, we have the following result: Theorem 1.2. ([18]) Let ( X , d) be a complete metric space and T : X → C B ( X ). If there exists 0 r < 1 such that T satisfies the condition
ψ(r )d(x, T x) d(x, y )
⇒
H ( T x, T y ) rM d (x, y )
for all x, y ∈ X , where ψ is given by (2), then T has a fixed point, that is, there exists a point z ∈ X such that z ∈ T z. The other basic notion for the development of our work is the concept of the partial metric space, that was introduced by Matthews [23] as a part of the study of denotational semantics of dataflow networks. He presented a modified version of the Banach contraction principle, more suitable in this context, see also [15,25]. In fact, the (complete) partial metric spaces constitute a suitable framework to model several distinguished examples of the theory of computation and also to model metric spaces via domain theory, see [12,16,17,19,23,28,30,33]. In this direction, Aydi et al. [4] introduced the concept of a partial Hausdorff metric and extended Nadler’s fixed point theorem in the setting of partial metric spaces. In view of the above considerations, the aim of this paper is to obtain a Kikkawa–Suzuki type fixed point theorem for multivalued mappings in a partial Hausdorff metric space. The presented results extend and unify some recently obtained comparable results for multivalued mappings (see [18] and the references therein). Moreover, using our results, we will prove the existence and uniqueness of the bounded solution of a functional equation arising in dynamic programming. Now on the letters R, R+ and N will denote the set of all real numbers, the set of all non-negative real numbers and the set of all positive integers, respectively. Consistent with [2–4,23], the following definitions and results will be needed in the sequel. Definition 1.3. ([23]) Let X be any non-empty set. A function p : X × X → R+ is said to be a partial metric if and only if for all x, y , z ∈ X the following conditions are satisfied: (P1) (P2) (P3) (P4)
p (x, x) = p ( y , y ) = p (x, y ) if and only if x = y; p (x, x) p (x, y ); p (x, y ) = p ( y , x); p (x, z) p (x, y ) + p ( y , z) − p ( y , y ).
The pair ( X , p ) is called a partial metric space. If p (x, y ) = 0, then (P1) and (P2) imply that x = y but the converse does not hold in general. A trivial example of a partial metric space is the pair (R+ , p ), where p : R+ × R+ → R+ is defined as p (x, y ) = max{x, y }, see also [1]. Example 1.4. ([23]) Let X = {[a, b]: a, b ∈ R , a b}. It is easy to show that the function p : X × X → R+ given by
p [a, b], [c , d] = max{b, d} − min{a, c } defines a partial metric on X . For further examples we refer to [2,9,28–30]. Note that each partial metric p on X generates a T 0 topology which has as a base, the family of the open balls (p-balls) { B p (x, ε ): x ∈ X , ε > 0} where
B p (x, ε ) = y ∈ X: p (x, y ) < p (x, x) + ε
τ p on X
for all x ∈ X and ε > 0. A sequence {xn }n∈ N in a partial metric space ( X , p ) is called convergent to a point x ∈ X , with respect to τ p , if and only if p (x, x) = limn→∞ p (x, xn ), see [23] for details. If p is a partial metric on X , then the function p S (x, y ) = 2p (x, y ) − p (x, x) − p ( y , y ) defines a metric on X . Further a sequence {xn }n∈ N converges in the metric space ( X , p S ) to a point x ∈ X if and only if
p (x, x) = lim p (x, xn ) = n→+∞
lim
n,m→+∞
p (xn , xm ).
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Definition 1.5. ([23]) Let ( X , p ) be a partial metric space. Then (a) A sequence {xn }n∈ N in X is called Cauchy if and only if limn,m→+∞ p (xn , xm ) exists and is finite. (b) A partial metric space ( X , p ) is said to be complete if every Cauchy sequence {xn }n∈ N in X converges with respect to τ p to a point x ∈ X such that p (x, x) = limn→+∞ p (x, xn ). Lemma 1.6. ([2,23]) Let ( X , p ) be a partial metric space. Then (a) A sequence {xn }n∈ N in X is Cauchy in ( X , p ) if and only if it is Cauchy in ( X , p S ). (b) A partial metric space ( X , p ) is complete if and only if the metric space ( X , p S ) is complete. Consistent with [4], let ( X , p ) be a partial metric space and let C B p ( X ) be the family of all non-empty, closed and bounded subsets of the partial metric space ( X , p ), induced by the partial metric p. Note that the closedness is taken from ( X , τ p ) (τ p is the topology induced by p) and the boundedness is given as follows: A is a bounded subset in ( X , p ) if there exist x0 ∈ X and M 0 such that for all a ∈ A, we have a ∈ B p (x0 , M ), that is, p (x0 , a) < p (x0 , x0 ) + M. For A , B ∈ C B p ( X ), x ∈ X , δ p : C B p ( X ) × C B p ( X ) → R+ define
p (x, A ) = inf p (x, a): a ∈ A ,
p ( A , B ) = inf p (x, y ): x ∈ A , y ∈ B ,
δ p ( A , B ) = sup p (a, B ): a ∈ A ,
δ p ( B , A ) = sup p (b, A ): b ∈ B ,
H p ( A , B ) = max δ p ( A , B ), δ p ( B , A ) .
It is easy to show that p (x, A ) = 0 implies that p S (x, A ) = 0, where
p S (x, A ) = inf p S (x, a): a ∈ A . Lemma 1.7. ([2]) Let ( X , p ) be a partial metric space and A be any non-empty subset of X , then a ∈ A if and only if p (a, A ) = p (a, a). Proposition 1.8. ([4]) Let ( X , p ) be a partial metric space. For any A , B , C ∈ C B p ( X ), we have the following: (i) (ii) (iii) (iv)
δ p ( A , A ) = sup{ p (a, a): a ∈ A }; δ p ( A , A ) δ p ( A , B ); δ p ( A , B ) = 0 ⇒ A ⊆ B; δ p ( A , B ) δ p ( A , C ) + δ p (C , B ) − infc∈C p (c , c ).
Proposition 1.9. ([4]) Let ( X , p ) be a partial metric space. For any A , B , C ∈ C B p ( X ), we have the following: (h1) (h2) (h3) (h4)
H p ( A , A ) H p ( A , B ); H p ( A , B ) = H p ( B , A ); H p ( A , B ) H p ( A , C ) + H p (C , B ) − infc ∈C p (c , c ); H p ( A , B ) = 0 ⇒ A = B.
The mapping H p : C B p ( X ) × C B p ( X ) → R+ is called the partial Hausdorff metric induced by p. Every Hausdorff metric is a partial Hausdorff metric but the converse is not true, see Example 2.6 in [4]. Lemma 1.10. ([4]) Let ( X , p ) be a partial metric space, A , B ∈ C B p ( X ) and h > 1, then for any a ∈ A, there exists b(a) ∈ B such that p (a, b(a)) hH p ( A , B ). Theorem 1.11. ([4]) Let ( X , p ) be a partial metric space. If T : X → C B p ( X ) is a multivalued mapping such that for all x, y ∈ X , we have H p ( T x, T y ) kp (x, y ), where k ∈ (0, 1), then T has a fixed point, that is, there exists a point z ∈ X such that z ∈ T z. 2. Fixed point results in partial Hausdorff metric spaces Let p : X × X → R+ be a partial metric and T : X → C B p ( X ) be a multivalued mapping. We define
M p (x, y ) = max p (x, y ), p (x, T x), p ( y , T y ),
p (x, T y ) + p ( y , T x) 2
.
Now we state and prove our main result. Theorem 2.1. Let ( X , p ) be a complete partial metric space, T : X → C B p ( X ) be a multivalued mapping and ψ : [0, 1) → (0, 1] be the non-increasing function defined by (2). If there exists 0 r < 1 such that T satisfies the condition
ψ(r ) p (x, T x) p (x, y )
⇒
H p ( T x, T y ) rM p (x, y )
(3)
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for all x, y ∈ X , then T has a fixed point, that is, there exists a point z ∈ X such that z ∈ T z. Proof. Let r1 be a real number such that 0 r < r1 < 1 and u 1 ∈ X . As T u 1 is non-empty, we can choose u 2 ∈ T u 1 . Clearly, if u 2 = u 1 the proof is finished and so we assume u 2 = u 1 . From Lemma 1.10, with h = √1r , there exists u 3 ∈ T u 2 such that 1
p (u 2 , u 3 ) √1r H p ( T u 1 , T u 2 ). Again, if u 3 = u 2 the proof is finished and so we assume u 3 = u 2 . Since ψ(r ) 1, we have 1
ψ(r ) p (u 1 , T u 1 ) p (u 1 , T u 1 ) p (u 1 , u 2 ). Now from (3), we obtain 1 p (u 2 , u 3 ) √ H p ( T u 1 , T u 2 ) r1
Thus
p (u 2 , u 3 )
√
r1 max p (u 1 , u 2 ), p (u 1 , T u 1 ), p (u 2 , T u 2 ),
√
r1 max p (u 1 , u 2 ), p (u 2 , u 3 ),
√
r1 max p (u 1 , u 2 ), p (u 2 , u 3 ),
√
p (u 1 , T u 2 ) + p (u 2 , T u 1 )
p (u 1 , u 3 ) + p (u 2 , u 2 ) 2 p (u 1 , u 2 ) + p (u 2 , u 3 ) 2
2
.
r1 max p (u 1 , u 2 ), p (u 2 , u 3 ) .
√
If max{ p (u 1 , u 2 ), p (u 2 , u 3 )} = p (u 2 , u 3 ), then p (u 2 , u 3 ) r1 p (u 2 , u 3 ) implies that p (u 2 , u 3 ) = 0 and we obtain p S (u 2 , u 3 ) 2p (u 2 , u 3 ) = 0 which further implies that p S (u 2 , u 3 ) = 0. Hence u 2 = u 3 ∈ T u 2 and the proof is finished. On the contrary if max{ p (u 1 , u 2 ), p (u 2 , u 3 )} = p (u 1 , u 2 ), then we have
p (u 2 , u 3 )
√
r1 p (u 1 , u 2 ).
Continuing this process, we can obtain a sequence {un } in X such that un+1 ∈ T un , un+1 = un and
√
p (un , un+1 ) ( r1 )n−1 p (u 1 , u 2 )
(4)
for every n > 1. This shows that limn→+∞ p (un , un+1 ) = 0. Since
p (un , un ) + p (un+1 , un+1 ) 2p (un , un+1 ), then we get
lim p (un , un ) = 0
n→+∞
Let
lim p (un+1 , un+1 ) = 0.
and
(5)
n→+∞
> 0 and pick N ∈ N large enough that for n N we have √ 1 2( r1 )n−1 √ p (u 1 , u 2 ) < . 1 − r1
Then, for every positive integer k > n N there is a m ∈ N such that k = n + m and we have
p S (un , un+m ) 2p (un , un+m ) 2 p (un , un+1 ) + · · · + p (un+m−1 , un+m )
√ √ √ 2 ( r1 )n−1 p (u 1 , u 2 ) + · · · + ( r1 )n+m−2 p (u 1 , u 2 ) 2( r1 )n−1
1 1−
√
r1
p (u 1 , u 2 ) < .
It is immediate to deduce that {un } is a Cauchy sequence in ( X , p S ) and by Lemma 1.6 {un } is Cauchy in ( X , p ). Moreover, since ( X , p ) is complete, again by Lemma 1.6 we have the completeness of ( X , p S ). It follows that there exists z ∈ X such that limn→+∞ un = z in ( X , p S ). Therefore limn→+∞ p S (un , z) = 0 implies
p ( z, z) = lim p (un , z) = n→+∞
lim
m,n→+∞
p (un , um ).
Now, since {un } is a Cauchy sequence in ( X , p S ), then we have
lim
p S (u n , u m ) = 0
lim
2p (un , um ) − lim
m,n→+∞
and so m,n→+∞
m→+∞
p (um , um ) − lim p (un , un ) = 0.
It follows from (5) that
lim
m→+∞
p (um , um ) = lim p (un , un ) = 0 n→+∞
n→+∞
M. Abbas et al. / Topology and its Applications 160 (2013) 553–563
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which further implies that
lim
m,n→+∞
p (u n , u m ) = 0
and
p ( z, z) = lim p (un , z) = lim p (un , un ) = 0. n→∞
n→∞
Now, the inequalities
p ( z, T x) p ( z, un+1 ) + p (un+1 , T x) − p (un+1 , un+1 ) and
p (un+1 , T x) p (un+1 , un ) + p (un , z) + p ( z, T x) − p (un , un ) − p ( z, z) give us
lim p (un+1 , T x) = lim p (un , T x) = p ( z, T x).
n→+∞
n→+∞
We claim that
p ( z, T x) r max p ( z, x), p (x, T x)
for all x = z. For if x = z, then p ( z, T x) r max{ p ( z, x), p (x, T x)} implies x ∈ T x and hence the proof is finished. Since p ( z, z) = limn→+∞ p (un , z) = 0, then there exists a positive integer n0 ∈ N such that
p ( z, un )
1 3
p ( z, x)
for all n n0 and this implies un = x. Since un+1 ∈ T un , then we have
ψ(r ) p (un , T un ) p (un , T un ) p (un , un+1 ) p (un , z) + p ( z, un+1 ) = p ( z, x) −
1 3
2 3
p ( z, x)
p ( z, x) p ( z, x) − p (un , z) p (un , x) − p (un , un ) p (un , x).
Hence, for any n n0 we get ψ(r ) p (un , T un ) p (un , x). Now from (3), we obtain
p (un+1 , T x) H p ( T un , T x)
p (un , T x) + p (x, T un ) r max p (un , x), p (un , T un ), p (x, T x), 2 p (un , T x) + p (x, un+1 ) r max p (un , x), p (un , un+1 ), p (x, T x), 2 r max p (un , z) + p ( z, x) − p ( z, z), p (un , un+1 ), p (x, T x), p (un , z) + p ( z, T x) − p ( z, z) + p (x, z) + p ( z, un+1 ) − p ( z, z) 2
r max p (un , z) + p ( z, x), p (un , un+1 ), p (x, T x), On taking the limit as n → +∞, we get
p ( z, T x) r max p ( z, x), p (x, T x), If max{ p ( z, x), p (x, T x),
p ( z, T x) r
p ( z, T x)+ p (x, z) } 2
p ( z, T x) + p (x, z) 2
=
p ( z, T x) + p (x, z) 2
p ( z, T x)+ p (x, z) , 2
.
then we have
.
Since r < 1, it follows that
p ( z, T x)
r 2−r
p (x, z) < rp (x, z) r max p ( z, x), p (x, T x)
p (un , z) + p ( z, T x) + p (x, z) + p ( z, un+1 ) 2
.
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M. Abbas et al. / Topology and its Applications 160 (2013) 553–563
and hence
p ( z, T x) r max p ( z, x), p (x, T x)
(6) p ( z, T x)+ p (x, z)
for all x = z. On the other hand, if max{ p ( z, x), p (x, T x), } is equal to p (x, z) or p (z, T x), we can deduce easily 2 that (6) holds. / Tz = Now we shall prove that z ∈ T z. First, consider the case when 0 r < 1/2 and suppose, by contradiction, that z ∈ T z, as T z is closed. Hence by Lemma 1.7, p ( z, T z) = p ( z, z) = 0. Then we can choose a ∈ T z such that 2rp (a, z) < p ( z, T z). / T z imply a = z, then from (6) with x = a, we have Since a ∈ T z and z ∈
p ( z, T a) r max p ( z, a), p (a, T a) .
(7)
Since a ∈ T z and ψ(r ) p ( z, T z) p ( z, T z) p ( z, a), therefore from (3) we have
H p ( T z, T a) r max p ( z, a), p ( z, T z), p (a, T a),
r max p ( z, a), p (a, T a), that is,
p ( z, T a) + p (a, T z) 2
p ( z, T a) + p (a, a)
2
p ( z, a) + p (a, T a) r max p ( z, a), p (a, T a), , 2
H p ( T z, T a) r max p ( z, a), p (a, T a) .
(8)
Since a ∈ T z, then p (a, T a) H p ( T z, T a). Therefore from (8) we obtain
H p ( T z, T a) r max p ( z, a), H p ( T z, T a) . Since r < 1, it follows that
H p ( T z, T a) rp ( z, a).
(9)
From (9), we obtain
p (a, T a) rp ( z, a). Starting from (P4), using the fact that p ( T a, T z) = inf{ p (x, y ): x ∈ T a, y ∈ T z} inf{ p (x, a): x ∈ T a} = p ( T a, a) since a ∈ T z and by (6), (8) and (9) we have
p ( z, T z) p ( z, T a) + p ( T a, T z) p ( z, T a) + H p ( T z, T a) r max p ( z, a), p (a, T a) + rp ( z, a)
= rp ( z, a) + rp ( z, a) = 2rp ( z, a) < p ( z, T z) that is a contradiction and hence z ∈ T z. Now consider the case when 12 r < 1. First we prove that
H p ( T x, T z) r max p (x, z), p (x, T x), p ( z, T z),
p (x, T z) + p ( z, T x)
(10)
2
for all x ∈ X such that x = z. For each n ∈ N, there exists yn ∈ T x such that p ( z, yn ) < p ( z, T x) + we have
p (x, T x) p (x, yn ) p (x, z) + p ( z, yn ) − p ( z, z) < p (x, z) + p ( z, T x) +
1 n
1 p (x, z) n
and consequently
p (x, z).
Hence by (6) we get
p (x, T x) < p (x, z) + r max p ( z, x), p (x, T x) +
1 n
p (x, z).
If max{ p ( z, x), p (x, T x)} = p (x, z), then from (11) we obtain
p (x, T x) < p (x, z) + rp ( z, x) +
1 n
p (x, z) < (1 + r ) +
1 n
(11)
p (x, z).
1 This implies that (1+ p (x, T x) < [1 + (1+1r )n ] p (x, z) and since ψ(r ) = 1 − r, it follows that r)
ψ(r ) p (x, T x) = (1 − r ) p (x, T x)
1
(1 + r )
p (x, T x) < 1 +
1
(1 + r )n
p (x, z).
M. Abbas et al. / Topology and its Applications 160 (2013) 553–563
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On taking the limit as n → +∞ we obtain ψ(r ) p (x, T x) p (x, z). Then from (3) with y = z we get (10). If p (x, z) < p (x, T x), then from (11) we obtain
p (x, T x) p (x, z) + rp (x, T x) +
1 n
p (x, z)
and hence (1 − r ) p (x, T x) (1 + n1 ) p (x, z). On taking again the limit as n → +∞, we get (1 − r ) p (x, T x) p (x, z), that is, ψ(r ) p (x, T x) p (x, z). Then from (3) with y = z we get (10). Since un+1 = un for each n ∈ N, we have un+1 = z or un = z and so the set J = {n: un = z} is infinite. From (10) with x = un , n ∈ J , we have
p ( u n +1 , T z ) H p ( T u n , T z )
p (u n , T z ) + p ( z , T u n )
r max p (un , z), p (un , T un ), p ( z, T z), 2 p ( u n , T z ) + p ( z , u n +1 ) . r max p (un , z), p (un , un+1 ), p ( z, T z), 2
On taking the limit as n → +∞, n ∈ J , we obtain
p ( z, T z) rp ( z, T z). Since r < 1, therefore p ( z, T z) = 0 = p ( z, z) and hence by Lemma 1.7 we have z ∈ T z, that is, z is a fixed point of T .
2
From Theorem 2.1 we deduce the following corollaries. Corollary 2.2. Let ( X , p ) be a complete partial metric space, T : X → C B p ( X ) be a multivalued mapping and ψ : [0, 1) → (0, 1] be the non-increasing function defined by (2). If there exists 0 r < 1 such that T satisfies the condition
ψ(r ) p (x, T x) p (x, y )
⇒
H p ( T x, T y ) r max p (x, y ), p (x, T x), p ( y , T y )
(12)
for all x, y ∈ X , then T has a fixed point, that is, there exists a point z ∈ X such that z ∈ T z. Corollary 2.3. Let ( X , p ) be a complete partial metric space, T : X → C B p ( X ) be a multivalued mapping and ψ : [0, 1) → (0, 1] be the non-increasing function defined by (2). If there exists 0 r < 1 such that T satisfies the condition
ψ(r ) p (x, T x) p (x, y )
⇒
H p ( T x, T y )
r
3
p (x, y ) + p (x, T x) + p ( y , T y )
(13)
for all x, y ∈ X , then T has a fixed point, that is, there exists a point z ∈ X such that z ∈ T z. In the case of single-valued mappings, Theorem 2.1 reduces to the following important corollary that will be used in the last section to prove the existence and uniqueness of the bounded solution of a functional equation arising in dynamic programming. Corollary 2.4. Let ( X , p ) be a complete partial metric space, T : X → X be a single-valued mapping and ψ : [0, 1) → (0, 1] be the non-increasing function defined by (2). If there exists 0 r < 1 such that T satisfies
ψ(r ) p (x, T x) p (x, y )
⇒
p ( T x, T y ) rM p (x, y )
for all x, y ∈ X , then T has a unique fixed point, that is, there exists a unique point z ∈ X such that z = T z. Proof. The existence of the fixed point follows from Theorem 2.1. To prove the uniqueness, assume that there exist z1 , z2 ∈ X with z1 = z2 such that z1 = T z1 and z2 = T z2 . Then
ψ(r ) p ( z1 , T z1 ) p ( z1 , T z1 ) = p ( z1 , z1 ) p ( z1 , z2 ) that implies
p ( z1 , z2 ) = p ( T z1 , T z2 )
p ( z2 , T z1 ) + p ( z1 , T z2 )
r max p ( z1 , z2 ), p ( z1 , T z1 ), p ( z2 , T z2 ), 2 r max p ( z1 , z2 ), p ( z1 , z1 ), p ( z2 , z2 ) rp ( z1 , z2 ).
We deduce that p ( z1 , z2 ) = 0, which further implies that p S ( z1 , z2 ) 2p ( z1 , z2 ) = 0 and hence z1 = z2 .
2
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Example 2.5. Let X = {0, 1, 2} and p : X × X → R+ be defined by
p (0, 0) = p (1, 1) = 0 and p (0, 1) = p (1, 0) =
1 4
p (2, 2) =
1 3
,
p (0, 2) = p (2, 0) =
,
2 5
p (1, 2) = p (2, 1) =
,
11 15
.
Then p is a partial metric on X . Let ψ(r ) be given by (2) and define T : X → C B p ( X ) by
Tx =
if x = 2, { 0} {0, 1} if x = 2.
Therefore, we get
2
max p (x, T x): x ∈ X =
min p (x, y ): x, y ∈ X and x = y =
and
5
Note that for any r 16 , we have ψ(r )
1 3
.
5 6
and then
5 6
and so ψ(r ) = 16 . Consequently we have
ψ(r ) p (x, T x) p (x, y ) holds for all x, y ∈ X with x = y. Put r =
H p ( T 0, T 1) = p (0, 0) = 0 rM p (0, 1), H p ( T 0, T 2) = p (0, 1) = H p ( T 1, T 2) = p (0, 1) =
1 4 1 4
< <
1
= rp (0, 2) rM p (0, 2),
3 11 18
= rp (1, 2) rM p (1, 2)
and similarly H p ( T x, T y ) rM p (x, y ) also holds true for x = y. Hence, for all x, y ∈ X , we get
ψ(r ) p (x, T x) p (x, y )
⇒
H p ( T x, T y ) rM p (x, y ).
Thus all the conditions of Theorem 2.1 are satisfied and x = 0 is the only fixed point of T . On the other hand, the metric p S induced by the partial metric p is given by
p S (0, 0) = p S (1, 1) = p S (2, 2) = 0, p S (1, 2) = p S (2, 1) =
17 15
p S (0, 1) = p S (1, 0) =
p S (0, 2) = p S (2, 0) =
,
7 15
1 2
,
.
Now it is easy to show that Theorem 2.1 is not applicable in this case. Since
ψ(r ) p S (0, T 0) = ψ(r ) p S (0, 0) = 0 p S (0, y ) is satisfied for any 0 r < 1 and y ∈ X , therefore for y = 2 we must have
H p S ( T 0, T 2) rM p S (0, 2). Now, we get
H p S ( T 0, T 2) = H p S {0}, {0, 1} =
1 2
and
M p S (0, 2) = max p S (0, 2), p S (0, T 0), p S (2, T 2),
= max
7 15
, 0,
7 15
Thus, for any 0 r < 1 we have
H p S ( T 0, T 2) rM p S (0, 2).
,
0+ 2
7 15
=
7 15
1
< . 2
p S (0, T 2) + p S (2, T 0) 2
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3. An application Generally, a dynamical process consists of a state space and a decision space. The state space is the set of the initial state, actions and transition model of the process; the decision space is the set of possible actions that are allowed for the process. In this section we assume that U and V are Banach spaces, W ⊆ U is a state space and D ⊆ V is a decision space. It is well known that the dynamic programming provides useful tools for mathematical optimization and computer programming as well. In particular, the problem of dynamic programming related to multistage process reduces to the problem of solving the functional equation
q(x) = sup f (x, y ) + G x, y , q
y∈ D
,
which further can be reformulated as
q(x) = sup g (x, y ) + G x, y , q y∈ D
x∈ W,
τ (x, y )
− b,
τ (x, y )
x∈ W,
(14)
where τ : W × D → W , f , g : W × D → R, G : W × D × R → R and b > 0. However, for the detailed background of the problem, the reader can refer to [5–8,26,32]. Here, we study the existence and uniqueness of the bounded solution of the functional equation (14). Let B ( W ) denote the set of all bounded real-valued functions on W and, for an arbitrary h ∈ B ( W ), define h = supx∈ W |h(x)|. Clearly, ( B ( W ), · ) endowed with the metric d defined by
d(h, k) = sup h(x) − k(x) x∈ W
for all h, k ∈ B ( W ), is a Banach space. Indeed, the convergence in the space B ( W ) with respect to · is uniform. Thus, if we consider a Cauchy sequence {hn } in B ( W ), then {hn } converges uniformly to a function, say h∗ , that is bounded and so h ∗ ∈ B ( W ). Moreover, we can define a partial metric p B by
p B (h, k) = d(h, k) + b
(15)
for all h, k ∈ B ( W ) and b > 0. We also define T : B ( W ) → B ( W ) by
T (h)(x) = sup g (x, y ) + G x, y , h
y∈ D
τ (x, y )
− b,
(16)
for all h ∈ B ( W ) and x ∈ W . Obviously, if the functions g and G are bounded then T is well-defined. Finally, let
p B (h, T (k)) + p B (k, T (h))
M p B (h, k) = max p B (h, k), p B h, T (h) , p B k, T (k) ,
2
.
We will prove the following theorem. Theorem 3.1. Assume that there exists 0 r < 1 such that
G x, y , h(x) − G x, y , k(x) rM p (h, k) B
where x ∈ W , y ∈ D, T : B ( W ) → B ( W ) is given by (16), the functions G : W × D × R → R and g : W × D → R are bounded. Then the functional equation (14) has a unique bounded solution. Proof. Since ( B ( W ), d) is complete and p BS (h, k) = 2d(h, k) for all h, k ∈ B ( W ) and x ∈ W , by Lemma 1.6 we deduce that ( B ( W ), p B ) is a complete partial metric space. Let λ be an arbitrary positive number, x ∈ W and h1 , h2 ∈ B ( W ), then there exist y 1 , y 2 ∈ D such that
T (h1 )(x) < g (x, y 1 ) + G x, y 1 , h1
τ (x, y 1 ) − b + λ,
T (h2 )(x) < g (x, y 2 ) + G x, y 2 , h2 τ (x, y 2 ) − b + λ, T (h1 )(x) g (x, y 2 ) + G x, y 2 , h1 τ (x, y 2 ) , T (h2 )(x) g (x, y 1 ) + G x, y 1 , h2 τ (x, y 1 ) . Then from (17) and (20), it follows easily that
(17) (18) (19) (20)
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T (h1 )(x) − T (h2 )(x) < G x, y 1 , h1
τ (x, y 1 ) − G x, y 1 , h2 τ (x, y 1 ) − b + λ
G x, y 1 , h1 τ (x, y 1 ) − G x, y 1 , h2 τ (x, y 1 ) − b + λ rM p B (h1 , h2 ) − b + λ.
Hence we get
T (h1 )(x) − T (h2 )(x) < rM p B (h1 , h2 ) − b + λ.
(21)
Similarly, from (18) and (19) we obtain
T (h2 )(x) − T (h1 )(x) < rM p B (h1 , h2 ) − b + λ.
(22)
Therefore, from (21) and (22) we have
T (h1 )(x) − T (h2 )(x) < rM p (h1 , h2 ) − b + λ, B
that is,
(23)
p B T (h1 ), T (h2 ) < rM p B (h1 , h2 ) + λ. Since the above inequality does not depend on x ∈ W and λ > 0 is taken arbitrary, then we conclude immediately that
p B T (h1 ), T (h2 ) rM p B (h1 , h2 ). In particular, the last inequality holds for any x ∈ W such that ψ(r ) p B (h, T (h)) p B (h, k) and so Corollary 2.4 is applicable in this case. Consequently, the mapping T has a unique fixed point, that is, the functional equation (14) has a unique bounded solution. 2 Acknowledgements The authors would like to thank the editor and reviewers for their helpful comments. The third author is supported by Università degli Studi di Palermo, Local University Project R.S. ex 60%. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28]
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